answer：First, I need to determine the total cost Samantha will incur for the tax services. The company charges a flat fee of 50 for preparing the tax documents.Next, there's an additional service fee of 5% of the tax refund amount. Samantha expects her tax refund to be 1,200.To calculate the service fee, I'll multiply 5% by 1,200, which equals 60.Finally, I'll add the flat fee and the service fee together to find the total cost: 50 plus 60 equals 110.

answer：First, I need to determine how much Alex spends each time he visits the food truck. The meal combo costs 8.50 and the fresh juice costs 3.00. Adding these together gives a total of 11.50 per visit.Next, since Alex visits the food truck 5 times a week, I'll multiply the cost per visit by the number of visits. So, 11.50 multiplied by 5 equals 57.50.Therefore, Alex spends a total of 57.50 on meals and drinks from the food truck each week.

answer：First, I need to determine how many bags Dr. Green's team has already produced in 14 days. Since they produce 500 bags each day, the total production after 14 days is 500 multiplied by 14, which equals 7,000 bags.Next, I'll calculate how many more bags are needed to reach the monthly target of 10,000 bags. Subtracting the already produced 7,000 bags from the target gives us 3,000 bags remaining.Finally, to find out how many additional days are required to produce the remaining 3,000 bags at a rate of 500 bags per day, I'll divide 3,000 by 500. This results in 6 more days needed to meet the monthly goal.

answer：Okay, so I have this problem about a retired footballer analyzing the trajectory of a football. It's divided into two parts. The first part is about calculating the trajectory without air resistance, and the second part incorporates the Magnus effect due to spin. Let me try to work through each part step by step.Starting with part 1: Trajectory Calculation.Alright, the football is kicked from ground level, which is the origin (0,0). It's given an initial velocity ( v_0 ) at an angle ( theta ) from the horizontal. I need to derive the parametric equations for the trajectory, find the maximum height ( H ), and the horizontal range ( R ).First, I remember that in projectile motion without air resistance, the only acceleration is due to gravity, acting downward. So, the horizontal and vertical motions can be treated separately.Let me denote the horizontal direction as x-axis and vertical as y-axis.The initial velocity components are:- Horizontal: ( v_{0x} = v_0 cos theta )- Vertical: ( v_{0y} = v_0 sin theta )Since there's no air resistance, the horizontal velocity remains constant throughout the motion. The vertical motion is influenced by gravity, so the vertical velocity will change over time.The parametric equations for projectile motion are usually given by:- ( x(t) = v_{0x} t )- ( y(t) = v_{0y} t - frac{1}{2} g t^2 )So, substituting the initial velocities:- ( x(t) = v_0 cos theta cdot t )- ( y(t) = v_0 sin theta cdot t - frac{1}{2} g t^2 )That should be the parametric equations. Now, moving on to maximum height ( H ).Maximum height occurs when the vertical component of the velocity becomes zero. So, I can find the time ( t_H ) when ( v_y = 0 ).The vertical velocity as a function of time is:( v_y(t) = v_{0y} - g t )Setting ( v_y(t_H) = 0 ):( 0 = v_0 sin theta - g t_H )So, ( t_H = frac{v_0 sin theta}{g} )Now, plugging this time into the vertical position equation to find ( H ):( H = y(t_H) = v_0 sin theta cdot t_H - frac{1}{2} g t_H^2 )Substituting ( t_H ):( H = v_0 sin theta cdot left( frac{v_0 sin theta}{g} right) - frac{1}{2} g left( frac{v_0 sin theta}{g} right)^2 )Simplify:( H = frac{v_0^2 sin^2 theta}{g} - frac{1}{2} cdot frac{v_0^2 sin^2 theta}{g} )Which simplifies further to:( H = frac{v_0^2 sin^2 theta}{2g} )Okay, that seems right. Now, the horizontal distance ( R ) is the range, which is the distance traveled when the ball returns to the ground. That happens when ( y(t) = 0 ).So, set ( y(t) = 0 ):( 0 = v_0 sin theta cdot t - frac{1}{2} g t^2 )Factor out t:( 0 = t left( v_0 sin theta - frac{1}{2} g t right) )So, the solutions are ( t = 0 ) (the initial time) and ( t = frac{2 v_0 sin theta}{g} ). That's the total time of flight.Now, plug this time into the horizontal position equation:( R = x(t) = v_0 cos theta cdot t )Substitute ( t = frac{2 v_0 sin theta}{g} ):( R = v_0 cos theta cdot frac{2 v_0 sin theta}{g} )Simplify:( R = frac{2 v_0^2 sin theta cos theta}{g} )Using the double-angle identity ( sin 2theta = 2 sin theta cos theta ), this becomes:( R = frac{v_0^2 sin 2theta}{g} )Alright, so that's part 1 done. The parametric equations are ( x(t) = v_0 cos theta cdot t ) and ( y(t) = v_0 sin theta cdot t - frac{1}{2} g t^2 ). The maximum height is ( H = frac{v_0^2 sin^2 theta}{2g} ) and the range is ( R = frac{v_0^2 sin 2theta}{g} ).Moving on to part 2: Spin and Magnus Effect.This part is more complex. The Magnus effect adds a force perpendicular to the velocity due to the spin of the ball. The additional acceleration is given by ( a_m = k cdot omega cdot v ), where ( k ) is a constant, ( omega ) is angular velocity, and ( v ) is the instantaneous velocity.Given values: ( k = 0.1 , text{m}^2/text{s} ), ( omega = 10 , text{rad/s} ), ( v_0 = 20 , text{m/s} ), and ( theta = 30^circ ).I need to derive the modified differential equations of motion and determine the new trajectory.First, let me recall that the Magnus force is a lateral force, which means it acts perpendicular to the direction of motion. The direction of the force depends on the direction of spin. For a football, if it's spinning around its axis, the Magnus effect will cause a deflection either to the left or right, depending on the spin direction.But in this case, since the problem doesn't specify the direction of spin, I might need to make an assumption or perhaps treat it as a vector quantity.Wait, the acceleration is given as ( a_m = k cdot omega cdot v ). Hmm, but acceleration is a vector. So, actually, the Magnus acceleration vector is perpendicular to both the angular velocity vector and the velocity vector. So, it's a cross product.But in the problem statement, it's written as ( a_m = k cdot omega cdot v ). That might be a scalar magnitude, but since acceleration is a vector, we need to consider its direction.Alternatively, perhaps the problem is simplifying it to a scalar model, assuming that the Magnus effect adds a constant acceleration in a particular direction, say the y-direction, but that might not be accurate.Wait, let's think about it. The Magnus force is given by ( F_m = k cdot omega cdot v ), but actually, in reality, it's ( F_m = k cdot omega times v ), which is a cross product. So, the force is perpendicular to both the spin axis and the velocity vector.But in this problem, the acceleration is given as ( a_m = k cdot omega cdot v ). So, perhaps they are considering the magnitude, but the direction is important.Alternatively, maybe the problem is assuming that the spin is such that the Magnus effect adds a lateral acceleration in the y-direction, but I need to clarify.Wait, maybe I should model the Magnus acceleration as a vector. Let me denote the angular velocity vector ( omega ) as pointing in some direction, say, along the z-axis (spin around the vertical axis), and the velocity vector ( v ) is in the x-y plane.Then, the cross product ( omega times v ) would result in a vector in the direction perpendicular to both, which would be in the y-z plane or x-z plane, depending on the spin.Wait, actually, if the ball is spinning around the z-axis, then ( omega ) is along z. The velocity ( v ) is in the x-y plane. So, the cross product ( omega times v ) would be in the direction perpendicular to both, which is the z-axis direction? Wait, no.Wait, cross product of z-axis and x-y plane vector. Let me recall, if ( omega = omega_z hat{k} ) and ( v = v_x hat{i} + v_y hat{j} ), then ( omega times v = omega_z (v_x hat{i} + v_y hat{j}) times hat{k} ).Wait, cross product of ( hat{k} ) with ( hat{i} ) is ( hat{j} ), and cross product of ( hat{k} ) with ( hat{j} ) is ( -hat{i} ). So, ( omega times v = omega_z (v_x hat{j} - v_y hat{i}) ).So, the Magnus force would be in the direction of ( omega times v ), which is a combination of x and y components.But in the problem statement, the acceleration is given as ( a_m = k cdot omega cdot v ). So, perhaps they are considering the magnitude, but in reality, it's a vector.Alternatively, maybe the problem is assuming that the Magnus effect only affects the vertical motion, but that might not be accurate.Wait, perhaps the problem is oversimplified, considering that the Magnus acceleration is in the vertical direction. But I think that's not correct because the Magnus effect causes a lateral force, not vertical.Wait, actually, the Magnus effect can cause a vertical force if the spin is around the horizontal axis, but in this case, if the spin is around the vertical axis, it would cause a lateral force.Wait, maybe I need to clarify the direction of the spin. Since the football is kicked at an angle, the spin could be around the axis perpendicular to the direction of motion, which would cause a Magnus force either to the left or right.But without knowing the direction of spin, it's difficult to model. Maybe the problem assumes that the spin is such that the Magnus force is acting in the positive y-direction or something.Alternatively, perhaps the problem is considering the Magnus effect as an additional acceleration in the y-direction, but that might not be correct.Wait, perhaps I should model the Magnus acceleration as a vector. Let me try that.Let me denote the angular velocity vector ( omega ) as having components. But the problem doesn't specify the direction of spin, so maybe it's assumed to be around the vertical axis, which would cause a lateral force in the x-y plane.Wait, but in reality, when a football is kicked, the spin is usually around the axis perpendicular to the direction of motion, which would be the vertical axis if the ball is moving horizontally. So, in this case, if the ball is kicked at an angle, the spin axis would be perpendicular to the velocity vector.Wait, maybe that's complicating things. Let me try to think differently.Given that the acceleration is ( a_m = k cdot omega cdot v ), but as a vector, it's ( vec{a}_m = k cdot vec{omega} times vec{v} ). So, the direction is determined by the cross product.But since the problem gives ( a_m ) as a scalar, maybe it's considering the magnitude, but we need to figure out the direction.Alternatively, perhaps the problem is assuming that the Magnus effect adds a constant acceleration in the y-direction, but I'm not sure.Wait, maybe I should consider that the Magnus force is always acting perpendicular to the velocity vector, so it affects both x and y components.But in that case, the differential equations would become coupled and more complex.Alternatively, perhaps the problem is assuming that the Magnus effect only affects the vertical motion, but I think that's not accurate.Wait, maybe I should look up the standard Magnus effect equations.Wait, no, I should try to derive it.So, the Magnus force is given by ( vec{F}_m = k cdot vec{omega} times vec{v} ). Dividing by mass, the acceleration is ( vec{a}_m = frac{k}{m} cdot vec{omega} times vec{v} ). But the problem states ( a_m = k cdot omega cdot v ), which is scalar. So, perhaps they are considering the magnitude, but the direction is important.Alternatively, maybe the problem is assuming that the Magnus acceleration is in the vertical direction, so it only affects the y-component.Wait, but that might not be the case. Let me think.If the ball is moving with velocity ( vec{v} ) and spinning with angular velocity ( vec{omega} ), the Magnus force is ( vec{F}_m = k cdot vec{omega} times vec{v} ). So, the direction is perpendicular to both ( vec{omega} ) and ( vec{v} ).Assuming that the spin is such that ( vec{omega} ) is along the z-axis (i.e., the ball is spinning around its vertical axis), then ( vec{omega} = omega hat{k} ). The velocity vector ( vec{v} ) has components ( v_x hat{i} + v_y hat{j} ). So, the cross product ( vec{omega} times vec{v} ) is ( omega (v_x hat{j} - v_y hat{i}) ).Therefore, the Magnus force is ( vec{F}_m = k cdot omega (v_x hat{j} - v_y hat{i}) ). Dividing by mass, the acceleration is ( vec{a}_m = k cdot omega (v_x hat{j} - v_y hat{i}) ).Wait, but in the problem statement, it's given as ( a_m = k cdot omega cdot v ). So, perhaps they are considering the magnitude, but the direction is important.So, in terms of components, the Magnus acceleration has x and y components:( a_{mx} = -k cdot omega cdot v_y )( a_{my} = k cdot omega cdot v_x )But wait, in the cross product, it's ( vec{omega} times vec{v} = omega (v_x hat{j} - v_y hat{i}) ), so the x-component is ( -k cdot omega cdot v_y ) and the y-component is ( k cdot omega cdot v_x ).Therefore, the total acceleration in x and y directions are:( a_x = a_{gx} + a_{mx} = 0 + (-k cdot omega cdot v_y) = -k cdot omega cdot v_y )( a_y = a_{gy} + a_{my} = -g + k cdot omega cdot v_x )So, the differential equations become:( frac{d^2 x}{dt^2} = -k cdot omega cdot frac{dy}{dt} )( frac{d^2 y}{dt^2} = -g + k cdot omega cdot frac{dx}{dt} )But these are coupled second-order differential equations, which might be a bit tricky to solve.Alternatively, we can write them in terms of velocities:Let ( v_x = frac{dx}{dt} ), ( v_y = frac{dy}{dt} ).Then, the equations are:( frac{dv_x}{dt} = -k cdot omega cdot v_y )( frac{dv_y}{dt} = -g + k cdot omega cdot v_x )These are coupled first-order differential equations.Given the initial conditions: at ( t = 0 ), ( x = 0 ), ( y = 0 ), ( v_x = v_0 cos theta ), ( v_y = v_0 sin theta ).So, we have a system of ODEs:1. ( frac{dv_x}{dt} = -k omega v_y )2. ( frac{dv_y}{dt} = -g + k omega v_x )This is a system of linear ODEs, which can be solved using various methods, such as Laplace transforms or matrix methods.Alternatively, we can write this system in matrix form:( frac{d}{dt} begin{pmatrix} v_x v_y end{pmatrix} = begin{pmatrix} 0 & -k omega k omega & 0 end{pmatrix} begin{pmatrix} v_x v_y end{pmatrix} + begin{pmatrix} 0 -g end{pmatrix} )This is a nonhomogeneous linear system. The homogeneous part is:( frac{d}{dt} begin{pmatrix} v_x v_y end{pmatrix} = begin{pmatrix} 0 & -k omega k omega & 0 end{pmatrix} begin{pmatrix} v_x v_y end{pmatrix} )The eigenvalues of the coefficient matrix can be found by solving ( det(A - lambda I) = 0 ), where ( A = begin{pmatrix} 0 & -k omega k omega & 0 end{pmatrix} ).The characteristic equation is ( lambda^2 + (k omega)^2 = 0 ), so ( lambda = pm i k omega ). Therefore, the homogeneous solution will involve sine and cosine terms with frequency ( k omega ).The particular solution can be found by assuming a steady-state solution for the nonhomogeneous term. Since the nonhomogeneous term is constant (-g in the y-component), we can assume a particular solution where ( v_x ) and ( v_y ) are constants.Let me denote the particular solution as ( begin{pmatrix} v_{x_p} v_{y_p} end{pmatrix} ).Substituting into the ODEs:1. ( 0 = -k omega v_{y_p} ) => ( v_{y_p} = 0 )2. ( 0 = -g + k omega v_{x_p} ) => ( v_{x_p} = frac{g}{k omega} )So, the particular solution is ( begin{pmatrix} frac{g}{k omega} 0 end{pmatrix} ).Therefore, the general solution is the homogeneous solution plus the particular solution.The homogeneous solution can be written as:( begin{pmatrix} v_x v_y end{pmatrix}_h = C_1 begin{pmatrix} cos(k omega t) sin(k omega t) end{pmatrix} + C_2 begin{pmatrix} sin(k omega t) -cos(k omega t) end{pmatrix} )So, the general solution is:( begin{pmatrix} v_x v_y end{pmatrix} = C_1 begin{pmatrix} cos(k omega t) sin(k omega t) end{pmatrix} + C_2 begin{pmatrix} sin(k omega t) -cos(k omega t) end{pmatrix} + begin{pmatrix} frac{g}{k omega} 0 end{pmatrix} )Now, applying the initial conditions at ( t = 0 ):( v_x(0) = v_0 cos theta = C_1 cos(0) + C_2 sin(0) + frac{g}{k omega} )=> ( v_0 cos theta = C_1 + frac{g}{k omega} )=> ( C_1 = v_0 cos theta - frac{g}{k omega} )Similarly, ( v_y(0) = v_0 sin theta = C_1 sin(0) + C_2 (-cos(0)) + 0 )=> ( v_0 sin theta = -C_2 )=> ( C_2 = -v_0 sin theta )Therefore, the velocity components are:( v_x(t) = left( v_0 cos theta - frac{g}{k omega} right) cos(k omega t) - v_0 sin theta sin(k omega t) + frac{g}{k omega} )( v_y(t) = left( v_0 cos theta - frac{g}{k omega} right) sin(k omega t) + v_0 sin theta cos(k omega t) )Now, to find the position components ( x(t) ) and ( y(t) ), we need to integrate the velocity components.Starting with ( v_x(t) ):( x(t) = int_0^t v_x(t') dt' + x(0) )Since ( x(0) = 0 ), we have:( x(t) = int_0^t left[ left( v_0 cos theta - frac{g}{k omega} right) cos(k omega t') - v_0 sin theta sin(k omega t') + frac{g}{k omega} right] dt' )Let me split the integral into three parts:1. ( left( v_0 cos theta - frac{g}{k omega} right) int_0^t cos(k omega t') dt' )2. ( -v_0 sin theta int_0^t sin(k omega t') dt' )3. ( frac{g}{k omega} int_0^t dt' )Compute each integral:1. ( left( v_0 cos theta - frac{g}{k omega} right) cdot frac{sin(k omega t)}{k omega} )2. ( -v_0 sin theta cdot left( -frac{cos(k omega t) - 1}{k omega} right) )3. ( frac{g}{k omega} cdot t )Simplify each term:1. ( frac{v_0 cos theta - frac{g}{k omega}}{k omega} sin(k omega t) )2. ( frac{v_0 sin theta}{k omega} (cos(k omega t) - 1) )3. ( frac{g}{k omega} t )So, combining all terms:( x(t) = frac{v_0 cos theta - frac{g}{k omega}}{k omega} sin(k omega t) + frac{v_0 sin theta}{k omega} (cos(k omega t) - 1) + frac{g}{k omega} t )Similarly, for ( v_y(t) ):( y(t) = int_0^t v_y(t') dt' + y(0) )Since ( y(0) = 0 ), we have:( y(t) = int_0^t left[ left( v_0 cos theta - frac{g}{k omega} right) sin(k omega t') + v_0 sin theta cos(k omega t') right] dt' )Again, split the integral:1. ( left( v_0 cos theta - frac{g}{k omega} right) int_0^t sin(k omega t') dt' )2. ( v_0 sin theta int_0^t cos(k omega t') dt' )Compute each integral:1. ( left( v_0 cos theta - frac{g}{k omega} right) cdot left( -frac{cos(k omega t) - 1}{k omega} right) )2. ( v_0 sin theta cdot frac{sin(k omega t)}{k omega} )Simplify each term:1. ( -frac{v_0 cos theta - frac{g}{k omega}}{k omega} (cos(k omega t) - 1) )2. ( frac{v_0 sin theta}{k omega} sin(k omega t) )So, combining all terms:( y(t) = -frac{v_0 cos theta - frac{g}{k omega}}{k omega} (cos(k omega t) - 1) + frac{v_0 sin theta}{k omega} sin(k omega t) )Simplify further:( y(t) = frac{v_0 cos theta - frac{g}{k omega}}{k omega} (1 - cos(k omega t)) + frac{v_0 sin theta}{k omega} sin(k omega t) )Alright, so now we have expressions for ( x(t) ) and ( y(t) ) in terms of ( t ). These are the modified parametric equations incorporating the Magnus effect.Given the values ( k = 0.1 , text{m}^2/text{s} ), ( omega = 10 , text{rad/s} ), ( v_0 = 20 , text{m/s} ), and ( theta = 30^circ ), let me plug these into the equations.First, compute ( k omega = 0.1 times 10 = 1 , text{s}^{-1} ).So, ( k omega = 1 ), which simplifies the equations a bit.Now, compute ( frac{g}{k omega} = frac{9.8}{1} = 9.8 , text{m/s} ).So, the velocity components become:( v_x(t) = (v_0 cos theta - 9.8) cos(t) - v_0 sin theta sin(t) + 9.8 )( v_y(t) = (v_0 cos theta - 9.8) sin(t) + v_0 sin theta cos(t) )Similarly, the position components:( x(t) = frac{v_0 cos theta - 9.8}{1} sin(t) + frac{v_0 sin theta}{1} (cos(t) - 1) + 9.8 t )( y(t) = frac{v_0 cos theta - 9.8}{1} (1 - cos(t)) + frac{v_0 sin theta}{1} sin(t) )Simplify further:( x(t) = (v_0 cos theta - 9.8) sin(t) + v_0 sin theta (cos(t) - 1) + 9.8 t )( y(t) = (v_0 cos theta - 9.8) (1 - cos(t)) + v_0 sin theta sin(t) )Now, plug in ( v_0 = 20 , text{m/s} ) and ( theta = 30^circ ).Compute ( v_0 cos theta = 20 cos 30^circ = 20 times (sqrt{3}/2) ≈ 20 times 0.8660 ≈ 17.32 , text{m/s} )Compute ( v_0 sin theta = 20 sin 30^circ = 20 times 0.5 = 10 , text{m/s} )So, substituting these values:( x(t) = (17.32 - 9.8) sin(t) + 10 (cos(t) - 1) + 9.8 t )Simplify:( x(t) = 7.52 sin(t) + 10 cos(t) - 10 + 9.8 t )( y(t) = (17.32 - 9.8) (1 - cos(t)) + 10 sin(t) )Simplify:( y(t) = 7.52 (1 - cos(t)) + 10 sin(t) )So, the parametric equations with Magnus effect are:( x(t) = 7.52 sin(t) + 10 cos(t) - 10 + 9.8 t )( y(t) = 7.52 (1 - cos(t)) + 10 sin(t) )These equations describe the trajectory of the football under the influence of the Magnus effect.To find the new trajectory path, we can plot these parametric equations, but since I can't plot here, I can describe the behavior.The x(t) equation has a linear term in t (9.8 t), which means the ball will continue moving forward indefinitely, but the y(t) equation will eventually cause the ball to come back down due to gravity. However, the Magnus effect introduces oscillations in both x and y due to the sine and cosine terms.But wait, actually, the presence of the 9.8 t term in x(t) suggests that the ball will have an increasing x-position over time, but the y(t) will eventually reach a maximum and then come back down. However, due to the Magnus effect, the x-position might oscillate around a linear trend.But in reality, the Magnus effect causes a curved trajectory, either to the left or right, depending on the spin. In this case, since the spin is causing a lateral force, the ball will curve in the direction determined by the spin.Given the positive coefficients in the x(t) equation, the ball will have a curved path, possibly to the right if the spin is such that the Magnus force is in the positive x-direction.But without knowing the exact direction of spin, it's hard to say, but the equations show that the x-position has a combination of oscillatory and linear motion, while the y-position has oscillatory motion around a parabolic path.However, since the Magnus effect is adding a force perpendicular to the velocity, the trajectory will be more complex than a simple parabola. It will likely be a curve that deviates from the original parabolic path.In summary, the modified differential equations are coupled and result in parametric equations that include both oscillatory and linear terms, leading to a curved trajectory.I think that's as far as I can go without actually solving the equations numerically. The key takeaway is that the Magnus effect introduces a lateral acceleration, causing the ball to curve, which is reflected in the parametric equations derived above.

answer：First, I'll calculate the total practice time for each day.On Monday, Alex practiced for 2 hours in the morning and 1.5 hours in the afternoon. Adding these together gives a total of 3.5 hours for Monday.Next, on Wednesday, he practiced for 1 hour in the morning and 2.5 hours in the afternoon. Adding these times results in 3.5 hours for Wednesday.Finally, on Saturday, Alex spent 3 hours practicing in the morning and 2 hours in the afternoon. Adding these gives a total of 5 hours for Saturday.To find the total practice time for the week, I'll add the totals from each day: 3.5 hours (Monday) + 3.5 hours (Wednesday) + 5 hours (Saturday) = 12 hours.

answer：First, I will list the number of kicks delivered in each of the five matches: 24, 28, 32, 36, and 40.Next, I will add these numbers together to find the total number of kicks.Adding 24 and 28 gives 52.Adding 32 to 52 results in 84.Adding 36 to 84 gives 120.Finally, adding 40 to 120 results in a total of 160 kicks.Therefore, the champion delivered a total of 160 kicks throughout the tournament.